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Chapter 16 Random Variables
math2200
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Life insurance
• Insurance company: a “death and disability” policy– Pay $10,000 when the client dies– Pay $5,000 if the client is permanently
disabled– Charge $50 per year
• Why $50?– Using actuarial information, the company can
calculate the expected value of the policy.
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Random variable
• The amount the company pays out on an individual policy is called a random variable.
• A random variable assumes a value based on the outcome of a random event. – Random variable is often denoted by a capital
letter, e.g., X– A particular value that it can have is often den
oted by the corresponding lower case letter, e.g., x
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Random variable
• Discrete– We can list all the outcomes (finite or countabl
e)– E.g. the amount the insurance pays out is eith
er $10,000, $5,000 or $0
• Continuous– any numeric value within a range of values.
• Example: the time you spend from home to school
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Probability model
• The collection of all possible values and the probabilities that they occur is called the probability model for the random variable.
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Example
• Death rate in any year is 1 out of every 1000 people
• 2 out of 1000 suffer some kind of disability• Probability model
Policyholder outcome
Payment
(x)
Probability (Pr(X=x))
Death 10,000 1/1000
Disability 5,000 2/1000
Neither 0 997/1000
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What does the insurance company expect?
• Suppose it insures exactly 1000 people• In a year,
– 1 customer dies– 2 are disabled– The insurance company pays $10,000 +
$5,000*2 = $20,000– Payment per customer: $20,000/1000 = $20– Earnings per customer: $50– Profit : $30 per customer!
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Expected value
• $20 is the expected payment per customer
• E(X) = 20=(10000 * 1 + 5000 * 2 + 0*997) / 1000
=10000*(1/1000) + 5000*(2/1000) + 0*(997/1000)
• E(X) = Σx* P(X=x)– Center of the distribution– A parameter of the model
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Expected value
• Of particular interest is the value we expect a random variable to take on, notated μ (for population mean) or E(X) for expected value.
• The expected value of a (discrete) random variable can be found by summing the products of each possible value and the probability that it occurs:
• Note: Be sure that every possible outcome is included in the sum and verify that you have a valid probability model to start with.
E X x P x
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• Most of the time, the company makes $50 per customer
• But, with small probabilities, the company needs to pay a lot ($10000 or $5000)
• The variation is big
• How to measure the variation?
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Spread
• For data, we calculated the standard deviation by first computing the deviation from the mean and squaring it. We do that with discrete random variables as well.
• The variance for a random variable is:
• The standard deviation for a random variable is:
22 Var X x P x
SD X Var X
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Variance and standard deviation
Policyholder outcome
Payment (x) Probability Pr(X=x)
Deviation
Death 10,000 1/1000 (10000-20) = 9980
Disability 5,000 2/1000 5000-20 =4980
Neither 0 997/1000 0 -20 = -20
Variance = 99802 (1/1000)+49802 (2/1000)+(-20)2 (997/1000) = 149,600
Standard deviation = square root of variance
Var(X) = Σ[x-E(X)]2 * P(X=x)
SD(X) = $386.78
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Properties of expected value and standard deviation
• Shifting– E(X+c) = E(X) + c– Var(X+c) = Var(X)– Example: Consider everyone in a company receiving
a $5000 increase in salary.
• Scaling– E(aX) = aE(X)– Var(aX) = a2 Var(X)– Example: Consider everyone in a company receiving
a 10% increase in salary.
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Properties of expected value and standard deviation
• Additivity– E(X ± Y) = E(X) ± E(Y)– If X and Y are independent
• Var(X ± Y) = Var(X) + Var(Y)• SD(X+Y) is NOT SD(X)+SD(Y)
• Suppose the outcomes for two customers are independent, what is the variance for the total payment to these two customers?– Var(X+Y) = Var(X)+Var(Y) = 149600 + 149600 = 2992
00• If one customer is insured twice as much, the va
riance is– Var(2X) = 4Var(X) = 4*149600 = 598400– SD(2X) = 2SD(X)
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X+Y and 2X
• Random variables do not simply add up together!– X and Y have the same probability model– But they are not the same random variables– Can NOT be written as X + X
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Example
• Sell used Isuzu Trooper and purchase a new Honda motor scooter– Selling Isuzu for a mean of $6940 with a
standard deviation $250– Purchase a new scooter for a mean of $1413
with a standard deviation $11
• How much money do I expect to have after the transaction? What is the standard deviation?
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Combining Random Variables (The Bad News)
• It would be nice if we could go directly from models of each random variable to a model for their sum.
• But, the probability model for the sum of two random variables is not necessarily the same as the model we started with even when the variables are independent.
• Thus, even though expected values may add, the probability model itself is different.
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• When two independent continuous random variables have Normal models, so does their sum or difference.
• This fact will let us apply our knowledge of Normal probabilities to questions about the sum or difference of independent random variables.
Combining Random Variables (The Good News)
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Combining normal random variables
• Example: packaging stereos– Stage 1: packing
• Normal with mean 9min and sd 1.5min
– Stage 2: boxing• Normal with mean 6min and sd 1min
• What is the probability that packing two consecutive systems take over 20 minutes?
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• X1: time for packing the first system– mean=9, sd = 1.5
• X2: time for packing the second system• T=X1+X2: total time to pack two systems
– E(T) = E(X1)+E(X2) = 9+9=18– Var(T) = Var(X1)+Var(X2) = 1.52 + 1.52 (assuming ind
ependence)– T is Normal with mean 18 and sd 2.12
• z-score = (20-18)/2.12 = 0.94– P(T>20) = P(Z>0.94) = 0.1736
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• What percentage of the stereo systems take longer to pack than to box?– P: time for packing a system– B: time for boxing a system– D=P-B: difference in times to pack and box a
system– The questions is P(D>0)=?– Assuming P and B are independent
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• D is still Normal• E(D) = E(P-B) = E(P)-E(B) = 9-6=3• Var(D) = Var(P-B) = Var(P)+Var(B) = 1.52 + 12 =
3.25• SD(D) = 1.80• D is Normal with mean 3 and sd 1.80• P(D>0) = 0.9525• About 95% of all the stereo systems will require
more time for packing than for boxing
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• If X is a random variable with expected value E(X)=µ and Y is a random variable with expected value E(Y)=ν, then the covariance of X and Y is defined as
Cov(X,Y)=E((X-µ)(Y- ν))
• The covariance measures how X and Y vary together.
Correlation and Covariance
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Some properties of covariance
• Cov(X,Y)=Cov(Y,X)
• Cov(X,X)=Var(X)
• Cov(cX,dY)=c*dCov(X,Y)
• Cov(X,Y) = E(XY)- µν
• If X and Y are independent, Cov(X,Y)=0– The converse is NOT true
• Var(X ± Y) = Var(X) + Var(Y) ± 2Cov(X,Y)
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• Covariance, unlike correlation, doesn’t have to be between -1 and 1.
• To fix the “problem” we can divide the covariance by each of the standard deviations to get the correlation:
Correlation and Covariance (cont.)
( , )( , )
X Y
Cov X YCorr X Y
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What Can Go Wrong?
• Don’t assume everything’s Normal.– You must Think about whether the Normality
Assumption is justified.
• Watch out for variables that aren’t independent:– You can add expected values for any two
random variables, but – you can only add variances of independent
random variables.
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What Can Go Wrong? (cont.)
• Don’t forget: Variances of independent random variables add. Standard deviations don’t.
• Don’t forget: Variances of independent random variables add, even when you’re looking at the difference between them.
• Don’t write independent instances of a random variable with notation that looks like they are the same variables.